Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{5a - 5}{a^2 + 4a - 5} \div \dfrac{5a - 45}{-3a^2 - 9a + 30} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{5a - 5}{a^2 + 4a - 5} \times \dfrac{-3a^2 - 9a + 30}{5a - 45} $ First factor out any common factors. $q = \dfrac{5(a - 1)}{a^2 + 4a - 5} \times \dfrac{-3(a^2 + 3a - 10)}{5(a - 9)} $ Then factor the quadratic expressions. $q = \dfrac {5(a - 1)} {(a + 5)(a - 1)} \times \dfrac {-3(a + 5)(a - 2)} {5(a - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {5(a - 1) \times -3(a + 5)(a - 2) } { (a + 5)(a - 1) \times 5(a - 9)} $ $q = \dfrac {-15(a + 5)(a - 2)(a - 1)} {5(a + 5)(a - 1)(a - 9)} $ Notice that $(a + 5)$ and $(a - 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-15\cancel{(a + 5)}(a - 2)(a - 1)} {5\cancel{(a + 5)}(a - 1)(a - 9)} $ We are dividing by $a + 5$ , so $a + 5 \neq 0$ Therefore, $a \neq -5$ $q = \dfrac {-15\cancel{(a + 5)}(a - 2)\cancel{(a - 1)}} {5\cancel{(a + 5)}\cancel{(a - 1)}(a - 9)} $ We are dividing by $a - 1$ , so $a - 1 \neq 0$ Therefore, $a \neq 1$ $q = \dfrac {-15(a - 2)} {5(a - 9)} $ $ q = \dfrac{-3(a - 2)}{a - 9}; a \neq -5; a \neq 1 $